Greedy Conjecture for Strings of Length 4

Abstract

In this short note, we prove that the greedy conjecture for the shortest common superstring problem is true for strings of length 4.

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1 Introduction

In the shortest common superstring (SCS) problem one is given a set $\mathcal{S}=\{s_1, \ldots , s_n\}$ of $n$ strings and the goal is to find a shortest string $s$ such that each $s_i$ is a substring of $s$. This is a well-known problem having applications in such areas as genome assembly and data compression.

The problem is known to be NP-hard [10] (even if the input strings have length $3$ or if the alphabet is binary [2]) and APX-hard [12]. The fastest known exact solutions just reduce the problem to the Travelling salesman problem and have running time $(\sum _{i=1}^n|s_i|)^{O(1)}2^n$ [1, 6–8]. The currently best known approximation ratio is $2\frac{11}{23}$ [11]. Better upper bounds are known for special cases when input strings have bounded length [4, 5]. A recent survey of known results (both practical and theoretical) is given in [3].

The well known greedy conjecture states that the following extremely simple greedy algorithm has approximation ratio $2$ [14]: find two strings with longest mutual overlap and merge them into one string, repeat the process till only one string is left. This intriguing conjecture is open for more than $25$ years already. There is a partial progress however: it is known that the conjecture is true for some orders in which the input strings are merged by the greedy algorithm [9, 15].

In this short note, we consider another special case. We prove that the greedy conjecture is true if the input strings have length $4$. (While for strings of length $3$ the conjecture follows from the fact that the greedy algorithm achieves$2$-approximation of the compression measure [13].) We do this by a careful analysis of possible overlaps produced by the greedy algorithm.

2 Preliminaries

An overlap ${\text {ov}}(a,b)$ of two strings $a$ and $b$ is defined as the longest suffix of $a$ which is also a prefix of $b$.

Let $\mathcal{S}=\{s_1, \ldots , s_n\}$ be a set of pairwise different $4$-strings where by an $r$-string we denote just a string of length exactly $r$. Denote by $s^{{\text {opt}}}$ and $s^{{\text {gr}}}$ an optimal solution and a greedy solution for $\mathcal S$, respectively. Our goal is thus to show that

$$\begin{aligned} |s^{{\text {gr}}}| \le 2\cdot |s^{{\text {opt}}}| \, . \end{aligned}$$

(1)

For technical reasons, we assume in this paper that in case of ties the greedy algorithm prefers strings of the form aaaa for $\mathtt{a} \in \Sigma$.

Let $\pi =(\pi _1, \ldots , \pi _n)$ be a permutation of $\{1, \ldots , n\}$. By overlapping $n$ input strings in this particular order one gets a superstring of length

$$\begin{aligned} \sum \limits _{i=1}^{n}|s_i| - \sum _{i=1}^{n-1}|{\text {ov}}(s_{\pi _i}, s_{\pi _{i+1}})| \, . \end{aligned}$$

(2)

The second term in the expression above is called a compression of $\mathcal S$ with respect to $\pi$. Thus, an equivalent reformulation of SCS is the following: find an order of $n$ input strings that maximizes the compression. By $c^{{\text {opt}}}$ and $c^{{\text {gr}}}$ we denote the compression of the optimal solution $s^{{\text {opt}}}$ and the greedy solution $s^{{\text {gr}}}$, respectively. By combining (1) with (2) we get an equivalent reformulation of what we need to prove:

$$\begin{aligned} 4n-c^{{\text {gr}}}\le 2\cdot (4n-c^{{\text {opt}}}) \, . \end{aligned}$$

(3)

For a string $t$ of length at most $3$, let $\#^{{\text {opt}}}(t)$ and $\#^{{\text {gr}}}(t)$ be the number of overlaps that are equal to $t$ in $s^{{\text {opt}}}$ and $s^{{\text {gr}}}$, respectively. Similarly, let $\#^{{\text {opt}}}_i$ and $\#^{{\text {gr}}}_i$ be the number of overlaps of length exactly $i$. Then (3) is equivalent to

$$\begin{aligned} 4n - \#^{{\text {gr}}}_1 - 2\#^{{\text {gr}}}_2 - 3\#^{{\text {gr}}}_3 \le 2 \cdot (4n - \#^{{\text {opt}}}_1 - 2\#^{{\text {opt}}}_2 - 3\#^{{\text {opt}}}_3) \end{aligned}$$

(4)

or

$$\begin{aligned} 2\#^{{\text {opt}}}_1 + 4\#^{{\text {opt}}}_2 + 6\#^{{\text {opt}}}_3 \le 4n + \#^{{\text {gr}}}_1 + 2\#^{{\text {gr}}}_2 + 3\#^{{\text {gr}}}_3 \, . \end{aligned}$$

(5)

Since $\#^{{\text {opt}}}_1 + \#^{{\text {opt}}}_2 + \#^{{\text {opt}}}_3 \le n$ it suffices to prove that

$$\begin{aligned} 2 \#^{{\text {opt}}}_3 \le 3 \#^{{\text {gr}}}_3 + 2 \#^{{\text {gr}}}_2 + \#^{{\text {gr}}}_1 \, . \end{aligned}$$

(6)

Let $\mathcal{S}_3^{{\text {gr}}}$ be the set of strings at the point of time when the greedy algorithm already merged all pairs of strings whose overlap is $3$ and there is no more overlaps of length $3$ left. In the following lemma we show that the number of overlaps equal to a $3$-string $t$ in the greedy solution cannot be much smaller than that of the optimal solution.

Lemma 1

For any $3$-string $t$, $\#^{{\text {gr}}}(t) \ge \#^{{\text {opt}}}(t) - 1$. Moreover, if $\#^{{\text {gr}}}(t) = \#^{{\text {opt}}}(t) - 1$ then $\mathcal{S}_3^{{\text {gr}}}$ contains a string with prefix $t$ and suffix $t$.

Proof

Assume, for the sake of contradiction, that $\#^{{\text {gr}}}(t) \le \#^{{\text {opt}}}(t) - 2$. The optimal solution contains $\#^{{\text {opt}}}(t)$ overlaps equal to $t$ and hence among the input $n$ strings there are at least $\#^{{\text {opt}}}(t)$ strings whose prefix is $t$ and at least $\#^{{\text {opt}}}(t)$ strings whose suffix is $t$. Now consider the set $\mathcal{S}_3^{{\text {gr}}}$. Since $\#^{{\text {gr}}}(t) \le \#^{{\text {opt}}}(t) - 2$, we conclude that $\mathcal{S}_3^{{\text {gr}}}$ contains at least two strings whose suffix is $t$ and at least two strings whose prefix is $t$. Hence there are two different strings in this set whose overlap is $t$ which contradicts to the fact that there are no more $3$-overlaps.   $\square$

In the following the strings from $\mathcal{S}_3^{{\text {gr}}}$ are called blocks. For a $3$-string $t$, we say that a block is $t$ -bad if its suffix and its prefix are equal to $t$ and moreover $\#^{{\text {gr}}}(t) = \#^{{\text {opt}}}(t)-1$. We call a block bad if it is $t$-bad for a $3$-string $t$ and good otherwise. Let $\#_{{\text {bad}}}$ and $\#_{{\text {good}}}$ be the number of overlaps in all bad and good blocks, respectively. Then clearly $\#_{{\text {bad}}}+ \#_{{\text {good}}}= \#^{{\text {gr}}}_3$ (recall that all the overlaps inside the blocks have length 3).

Note that if there are no bad blocks then already Lemma 1 is sufficient to prove (6): in this case, $\#^{{\text {gr}}}(t) \ge \#^{{\text {opt}}}(t)$ and therefore $\#^{{\text {gr}}}_3 \ge \#^{{\text {opt}}}_3$.

Next, we consider bad blocks of fixed length: for a $3$-string $t$, let

$$\begin{aligned} \chi _{=i}(t) = [S_3^{gr} \text {contains a}\ t{-bad\;block\;of\;length \;exactly}\ i]. \end{aligned}$$

(throughout the paper, we use the standard Iverson brackets: $[P]$ is equal to $1$ if $P$ is true and is equal to $0$ otherwise). Further, let

$$ \chi _{=i} = \sum \limits _{|t|=3}\chi _{=i}(t) \, .$$

Functions $\chi _{>i}(t)$, $\chi _{\ge i} (t)$, $\chi _{>i}$, and $\chi _{\ge i}$ are defined in a similar fashion.

Note that $\chi _{=4} = 0$. Indeed a bad block of length $4$ must have a form aaaa. Also, $\#^{{\text {opt}}}(\mathtt{aaaa})>0$ and hence $\mathcal S$ contains another string starting or ending with aaa. But then the greedy algorithm must merge these two strings (as it prefers strings of the form aaaa). Hence for any $3$-string $t$, $\chi _{\ge 5}(t)$ is exactly the number of $t$-bad blocks.

Lemma 2

For any $3$-string $t$,

$$\begin{aligned} \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t) \} + \chi _{\ge 5}(t) = \#^{{\text {opt}}}(t) \, . \end{aligned}$$

Proof

Consider the following two cases:

• 1. $\#^{{\text {gr}}}(t) \ge \#^{{\text {opt}}}(t)$, then $\min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t)\} = \#^{{\text {opt}}}(t)$ and $\chi _{\ge 5} (t) = 0$.

• 2. $\#^{{\text {gr}}}(t) < \#^{{\text {opt}}}(t)$, then by Lemma 1, $\min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t)\} = \#^{{\text {opt}}}(t) - 1$. There is at least one block starting with $t$ and ending with $t$. Moreover there cannot be two different such blocks as otherwise the greedy algorithm would merge them. Therefore, there is exactly one $t$-bad block, i.e., $\chi _{\ge 5} (t) = 1$.    $\square$

By summing up the equality from Lemma 2 over all strings $t$ of length $3$ we get the following corollary.

Corollary 1

$$ \sum \limits _{|t|=3} \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t) \} + \chi _{\ge 5} = \#^{{\text {opt}}}_3 . $$

Assume now that $\chi _{=5}= 0$. Then due to the fact that a bad block of length exactly $i$ contains $i-4$ overlaps we have that $2 \chi _{>5}\le \#^{{\text {gr}}}_3$. By adding twice the $\sum \limits _{|t|=3} \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t) \}$ to both sides of this inequality and applying Corollary 1 we get

$$\begin{aligned} 2 \#^{{\text {opt}}}_3 \le \#^{{\text {gr}}}_3 + 2 \sum \limits _{|t|=3} \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t) \} \le 3 \#^{{\text {gr}}}_3 , \end{aligned}$$

which implies (6).

Hence the most tricky case is when there are bad blocks of length $5$. The rest of the paper is devoted to the analysis of this case. Note that such blocks have the form $\mathtt{ababa}$ (for different letters $\mathtt{a}, \mathtt{b} \in \Sigma$) and therefore these are aba-bad blocks. To analyze such blocks carefully we introduce the following definitions. For a $3$-string $t$ and $1 \le i \le 5$, $B_i(t)=0$ if either $t$ is not of the form aba, or $t$ is of the form aba and there is no block ababa. In the remaining case (i.e., $t$ is of the form aba and there is a block ababa) $B_i$’s are defined as follows:

$$\begin{aligned} B_1(\mathsf{aba })&= [\#^{{\text {gr}}}(\mathsf{bab }) > \#^{{\text {opt}}}(\mathsf{bab })] , \\ \nonumber B_2(\mathsf{aba })&= [\text {there exists a block with prefix}\ \mathsf{ba }\ \text {or suffix}\ \mathsf{ab }] , \\ \nonumber B_3(\mathsf{aba })&= [\text {there exists a block except}\ \mathsf{ababa }\ \text {with prefix}\ \mathsf{ab }\ \text {or suffix}\ \mathsf{ba }] , \\ \nonumber B_4(\mathsf{aba })&= [\text {there exists a good block of length at least 5} \\ \nonumber&\quad \; \; \;\text {containing}\ \mathsf{aba }\ \text {or}\ \mathsf{bab }\ \text {as a proper substring}] ,\\ \nonumber B_5(\mathsf{aba })&= [B_2(\mathsf{aba }) = 0\, \text {and}\, B_3(\mathsf{aba }) = 0\ \text {and there exists a bad block of}\\ \nonumber&\quad \; \; \text {length at least 7 containing}\ \mathsf{aba }\ \text {or}\ \mathsf{bab }\ \text {as substring} ]. \end{aligned}$$

Further, let for $1 \le i \le 5$, $B_i = \sum \limits _{|t| = 3}B_i(t)$.

Now we show $B_i$’s provide an upper bound for the number of bad blocks of length exactly $5$.

Lemma 3

$\chi _{=5} \le \sum \limits _{i=1}^5 B_i$.

Proof

Note that if $3$-string $t$ is not of the form aba then $\chi _{=5}(t)=0$ so the string $t$ contributes nothing to the left-hand side of the inequality. We now focus on $3$-strings $t$ of the form aba. It is sufficient to prove the following inequality:

$$\begin{aligned} \chi _{=5}(\mathtt{aba}) \le \sum \limits _{i=1}^{5} B_i(\mathtt{aba}) \end{aligned}$$

(7)

Assume that $\chi _{=5}(\mathtt{aba}) = 1$ and $B_1(\mathtt{aba}) = 0$ as otherwise the inequality holds for trivial reasons. From $B_1(\mathtt{aba}) = 0$ and Lemma 1 we have that $\#^{{\text {opt}}}(\mathtt{bab}) - 1 \le \#^{{\text {gr}}}(\mathtt{bab}) \le \#^{{\text {opt}}}(\mathtt{bab})$. Since $\#^{{\text {gr}}}(\mathtt{bab}) > 0$ (because $\mathcal{S}_3^{{\text {gr}}}$ contains the block $\mathtt{ababa}$ by definition of $\chi _{=5}(\mathtt{aba})$) we have that $\#^{{\text {opt}}}(\mathtt{bab}) > 0$, i.e. the optimal solution has at least one overlap of the form $\mathtt{bab}$. Depending of the location of this overlap in the optimal string we consider the following cases:

• 1. The overlap bab in the optimal solution is contained as a substring of $\mathtt{ababa}$. Since $\#^{{\text {opt}}}(\mathtt{aba}) > 0$, $\mathcal S$ contains at least one string except $\mathtt{abab}$ and $\mathtt{baba}$ containing $\mathtt{aba}$ as substring.

• 2. The overlap $\mathtt{bab}$ in the optimal solution is not in $\mathtt{ababa}$. Hence $\mathcal S$ contains at least one string except $\mathtt{abab}$ and $\mathtt{baba}$ containing $\mathtt{bab}$.

So in both cases there exists a string in $\mathcal S$ except $\mathtt{abab}$ and $\mathtt{baba}$ that contains $t' = \mathtt{aba}$ or $t' = \mathtt{bab}$. This string is contained by some block $r \in {\mathcal{S}_3^{{\text {gr}}}}$ and besides $r \ne \mathtt{ababa}$ and $r \ne \mathtt{babab}$. Consider the following cases:

• 1. $r$ is a good block. Then $B_4(\mathtt{aba}) > 0$ if $t'$ is a proper substring of $r$ and $B_2(\mathtt{aba}) + B_3(\mathtt{aba}) > 0$ otherwise. Therefore (7) holds.

• 2. $r$ is a bad block of length $5$. Then this block has a form $\mathtt{ababa}$ or $\mathtt{babab}$, a contradiction.

• 3. $r$ is a bad block of length $6$. If $t'$ is a prefix or a suffix of $r$ then $B_2(\mathtt{aba}) + B_3(\mathtt{aba}) > 0$. Otherwise either $r = \mathtt{r_1 t'_1 t'_2 t'_1 r_5 r_6}$ or $r = \mathtt{r_1 r_2 t'_1 t'_2 t'_1 r_6}$ where $t' = \mathtt{t'_1 t'_2 t'_3}$. Since $r$ is a bad block either $\mathtt{t'_1 t'_1 t'_2 t'_1 t'_1 t'_2}$ or $\mathtt{t'_2 t'_1 t'_1 t'_2 t'_1 t'_1}$. Finally, since either $t' = \mathtt{aba}$ or $t' = \mathtt{bab}$ in both these cases $r$ has a prefix or a suffix $\mathtt{ab}$ or $\mathtt{ba}$. Then $B_2(\mathtt{aba}) + B_3(\mathtt{aba}) > 0$ and (7) holds.

• 4. $r$ is a bad block of length at least $7$. Then $B_5(\mathtt{aba}) > 0$ and (7) holds.   $\square$

3 The Proof of the Main Theorem

In this section we prove the main result of this note: we first state auxiliary lemmas providing upper bounds on $B_i$’s, then show how these lemmas imply the main result of the paper, and then provide the proofs of all the lemmas.

Lemma 4

$B_1 + \sum \limits _{|t|=3} \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t) \} \le \#^{{\text {gr}}}_3$.

Lemma 5

$B_2 \le \#^{{\text {gr}}}_2$.

Lemma 6

$B_3 \le \#^{{\text {gr}}}_1 + \#^{{\text {gr}}}_2$.

Lemma 7

$B_4 \le \#_{{\text {good}}}$.

Lemma 8

$B_5 + 2 \chi _{>5}+ \chi _{=5}\le \#_{{\text {bad}}}$.

Theorem 1

The greedy algorithm for strings of length $4$ that prefers strings of the form aaaa in case of ties is $2$-approximate.

Proof

By adding the inequalities from Lemmas 5–8 to twice the inequality from Lemma 4 and applying equality $\#_{{\text {bad}}}+ \#_{{\text {good}}}= \#^{{\text {gr}}}_3$ one gets

$$\begin{aligned} 2 B_1 + B_2 + B_3 + B_4 + B_5 + 2 \chi _{>5}+ \chi _{=5}+ 2 \sum \limits _{|t|=3} \min \{&\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t) \} \le \\ \nonumber&3 \#^{{\text {gr}}}_3 + 2 \#^{{\text {gr}}}_2 + \#^{{\text {gr}}}_1 \, . \end{aligned}$$

By further adding the inequality from Lemma 3 we get

$$\begin{aligned} 2 \sum \limits _{|t|=3} \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t) \} + 2 \chi _{>5}+ 2 \chi _{=5}+ B_1 \le 3 \#^{{\text {gr}}}_3 + 2 \#^{{\text {gr}}}_2 + \#^{{\text {gr}}}_1 \, . \end{aligned}$$

Finally, applying Corollary 1 we get

$$\begin{aligned} 2 \#^{{\text {opt}}}_3 + B_1 \le 3 \#^{{\text {gr}}}_3 + 2 \#^{{\text {gr}}}_2 + \#^{{\text {gr}}}_1 \end{aligned}$$

which implies (6).    $\square$

Proof

(of Lemma 4 ). We have

$$\begin{aligned} B_1 + \sum \limits _{|t|=3} \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t) \} = \sum \limits _{|t| = 3} (B_1(t) + \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t)\})\\ \nonumber \; \,= \sum \limits _{t \ne \mathtt{aba}} (B_1(t) + \min \{\#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t)\}) + \sum \limits _{\mathtt{a}, \mathtt{b} \in \Sigma }(B_1(\mathtt{aba})+ B_1(\mathtt{bab}) \\ \nonumber + \min \{\#^{{\text {gr}}}(\mathtt{aba}), \#^{{\text {opt}}}(\mathtt{aba})\} \;+ \min \{\#^{{\text {gr}}}(\mathtt{bab}), \#^{{\text {opt}}}(\mathtt{bab})\}) \end{aligned}$$

To prove this lemma, we consider the following cases:

Case 1

If $t \ne \mathtt{aba}$ then $B_1(t) = 0$ and hence

$$\begin{aligned} B_1(t) + \min \{ \#^{{\text {gr}}}(t), \#^{{\text {opt}}}(t)\} \le \#^{{\text {gr}}}(t) \, . \end{aligned}$$

Case 2

If $t = \mathtt{aba}$ and $B_1(\mathtt{aba}) + B_1(\mathtt{bab}) = 0$ then

$$\begin{aligned} B_1(\mathtt{aba}) + B_1(\mathtt{bab}) +&\min \{\#^{{\text {gr}}}(\mathtt{aba}), \#^{{\text {opt}}}(\mathtt{aba})\}\\ \nonumber&+ \min \{\#^{{\text {gr}}}(\mathtt{bab}), \#^{{\text {opt}}}(\mathtt{bab}))\} \le \#^{{\text {gr}}}(\mathtt{aba}) + \#^{{\text {gr}}}(\mathtt{bab}) \end{aligned}$$

Case 3

If $t = \mathtt{aba}$ and $B_1(\mathtt{aba}) = 1$ then $B_1(\mathtt{bab}) = 0$ and, by definition of $B_1$, $\#^{{\text {gr}}}(\mathtt{bab}) > \#^{{\text {opt}}}(\mathtt{bab})$. Hence

$$\begin{aligned} \,B_1(\mathtt{aba}) + B_1(\mathtt{bab}) + \min&\{\#^{{\text {gr}}}(\mathtt{aba}), \#^{{\text {opt}}}(\mathtt{aba})\} + \min \{\#^{{\text {gr}}}(\mathtt{bab}), \#^{{\text {opt}}}(\mathtt{bab})\} \\ \nonumber = 1 + \min \{\#^{{\text {gr}}}&(\mathtt{aba}), \#^{{\text {opt}}}(\mathtt{aba})\} + \min \{\#^{{\text {gr}}}(\mathtt{bab}), \#^{{\text {opt}}}(\mathtt{bab})\}\\ \nonumber&\le 1 + \#^{{\text {gr}}}(\mathtt{aba}) + \#^{{\text {opt}}}(\mathtt{bab}) \\ \nonumber \le 1&+ \#^{{\text {gr}}}(\mathtt{aba}) + \#^{{\text {gr}}}(\mathtt{bab}) - 1 = \#^{{\text {gr}}}(\mathtt{aba}) + \#^{{\text {gr}}}(\mathtt{bab}) \end{aligned}$$

Case 4

If $t = \mathtt{aba}$ and $B_1(\mathtt{bab}) = 1$. This case is similar to Case 3.   $\square$

Proof

(of Lemma 5 ). We show that $B_2 \le \#^{{\text {gr}}}_2$. If $B_2(t)>0$ then $t$ is of the form aba and there exists a block with prefix ba or suffix ab. Since $B_2(t) > 0$ there exists a pair of blocks: $\mathtt{ababa}$ and a block with a $2$-prefix $\mathtt{ba}$ or a $2$-suffix $\mathtt{ab}$. Note that for different strings $t$ these pairs of blocks do not intersect and cannot be merged with $2$-overlaps because the sets $\{\mathtt{a}, \mathtt{b}\}$ are different. Note that at least one block in this pair must be merged with $2$-overlap with some block otherwise this pair of blocks must be merged by the greedy algorithm. Thus $\sum \limits _{t} B_2(t) < \#^{{\text {gr}}}_2$

    $\square$

For Lemma 6 we need the following auxiliary definitions. Let ${\text {Pref}}(\mathtt{a}, \mathtt{b}) = \emptyset$ if there is no block $\mathtt{ababa}$ and the set of blocks with prefix $\mathtt{ab}$ otherwise. Similarly, let ${\text {Suff}}(\mathtt{a}, \mathtt{b}) = \emptyset$ if there is no block $\mathtt{ababa}$ and the set of blocks with suffix $\mathtt{ba}$ otherwise. Then it is easy to see that:

$$\begin{aligned} (\mathtt{a} \ne \mathtt{a'} \vee \mathtt{b} \ne \mathtt{b'}) \Rightarrow ({\text {Pref}}(\mathtt{a}, \mathtt{b}) \cap {\text {Pref}}(\mathtt{a'}, \mathtt{b'}) = \emptyset \wedge {\text {Suff}}(\mathtt{a}, \mathtt{b}) \cap {\text {Suff}}(\mathtt{a'}, \mathtt{b'}) = \emptyset ) \, \end{aligned}$$

Let

$${\text {Pref}}(\mathtt{a}) = \bigcup \limits _{\mathtt{b} \in \Sigma } {\text {Pref}}(\mathtt{a}, \mathtt{b}) \text { and } {\text {Suff}}(\mathtt{a}) = \bigcup \limits _{\mathtt{b} \in \Sigma } {\text {Suff}}(\mathtt{a}, \mathtt{b}) \, .$$

Lemma 9

If $\mathtt{a} \ne \mathtt{c}$ then the set of $1$- and $2$-suffixes of strings from ${\text {Suff}}(\mathtt{a})$ does not intersect the set of $1$- and $2$-prefixes of strings from ${\text {Pref}}(\mathtt{c})$.

Proof

All $1$-suffixes of strings from ${\text {Suff}}(\mathtt{a})$ are equal to $\mathtt{a}$ while all $1$-prefixes of strings from ${\text {Pref}}(\mathtt{c})$ are equal to $\mathtt{c}$, hence they do not intersect.

Assume that $2$-suffix of $b_1 \in {\text {Suff}}(a)$ equals to $2$-prefix of block $b_2 \in {\text {Pref}}(c)$. $2$-suffix of block $b_1$ has the form $\mathtt{xa}$ and $2$-prefix of $b_2$ has the form $\mathtt{cy}$ so $x = \mathtt{c}, y = \mathtt{a}$. Hence $b_1$ has form $\mathtt{acaca}$ and $b_2$ has form $\mathtt{cacac}$, a contradiction.    $\square$

Proof

(of Lemma 6 ). $B_3(t) > 0$ only for $t = \mathtt{aba}$: $B_3 = \sum \limits _{t} B_3(t) = \sum \limits _{\mathtt{a}} \sum \limits _{\mathtt{b}} B_3(\mathtt{aba})$.

By Lemma 9 one can form sets $X_1^a$ of $1$-overlaps of strings from ${\text {Suff}}(a)$ and ${\text {Pref}}(a)$ counted in $\#^{{\text {gr}}}_1$. The lemma guarantees that these sets are disjoint. Similarly we can form sets $X_2^\mathtt{a}$ from $2$-overlaps of strings from ${\text {Suff}}(\mathtt{a})$ and ${\text {Pref}}(\mathtt{a})$. Hence

$$\begin{aligned} \sum \limits _{\mathtt{a}} |X_1^\mathtt{a}| \le \#^{{\text {gr}}}_1 \text { and } \sum \limits _{\mathtt{a}} |X_2^\mathtt{a}| \le \#^{{\text {gr}}}_2 \, . \end{aligned}$$

(8)

Since for each nonzero $\chi _{=5}(\mathtt{aba})$ there exists a block $\mathtt{ababa}$ we have, for each $\mathtt{a}$,

$$\begin{aligned} \sum \limits _{\mathtt{b}} B_3(\mathtt{aba}) \le \min \{|{\text {Pref}}(\mathtt{a})|, |{\text {Suff}}(\mathtt{a})|\} \, . \end{aligned}$$

(9)

Since for each block $\mathtt{ababa}$ with $B_3(\mathtt{aba}) > 0$ there exists by definition a string with prefix $\mathtt{ab}$ or suffix $\mathtt{ba}$, we have:

$$\begin{aligned} \sum \limits _{\mathtt{b}} B_3(\mathtt{aba}) < \max \{|{\text {Pref}}(\mathtt{a})|, |{\text {Suff}}(\mathtt{a})|\} \, . \end{aligned}$$

(10)

Assume that $|{\text {Pref}}(\mathtt{a})| \le |{\text {Suff}}(\mathtt{a})|$ (the opposite case is symmetric). Let us show that

$$\begin{aligned} |X_1^a| + |X_2^a| \ge \sum \limits _{b} B_3(\mathtt{aba}) \,. \end{aligned}$$

(11)

For this, assume the contrary. It follows from (9) and (10) that

$$ |X_1^\mathtt{a}| + |X_2^\mathtt{a}| \le |{\text {Pref}}(\mathtt{a})| - 1 \text { and } |X_1^\mathtt{a}| + |X_2^\mathtt{a}| \le |{\text {Suff}}(\mathtt{a})| - 2 \, .$$

Hence there exists at least one block from ${\text {Pref}}(\mathtt{a})$ whose prefix is not used in overlaps and there exist at least two blocks from ${\text {Suff}}(\mathtt{a})$ whose suffixes are not used in overlaps. But this prefix can be merged with one of these suffixes, a contradiction establishing (11).

Finally, by summing (11) for all $\mathtt{a}$ and applying (8) we get the required inequality:

$$ \sum \limits _{\mathtt{a} \in \Sigma } \sum \limits _{\mathtt{b} \in \Sigma } B_3(\mathtt{aba}) \le \sum _{\mathtt{a}} (|X_1^\mathtt{a}| + |X_2^\mathtt{a}|) \le \#^{{\text {gr}}}_1 + \#^{{\text {gr}}}_2 \, .$$

   $\square$

Proof

(of Lemma 7 ). If for some $\mathtt{a}, \mathtt{b} \in \Sigma$, $B_4(\mathtt{aba}) +B_4(\mathtt{bab}) = 1$, then either $\mathtt{aba}$ or $\mathtt{bab}$ is contained by a good block as a proper substring, so there exists at least one overlap by $t$ in a good block. Hence

$$B_4 = \sum \limits _{|t|=3} B_4(t) \le \#_{{\text {good}}}\, .$$

   $\square$

Proof

(of Lemma 8 ). Let $\#_{{\text {bad}}}^i$ be the number of overlaps in bad blocks of length $i$.

Let $B^i_5(t) = [ i \ge 7 \wedge t = \mathtt{aba} \wedge B_2(t) = B_3(t) = 0 \wedge \text {there exists a block} \; \mathtt{ababa} \text {and a bad-block of length i which contains}\ \mathtt{aba}\ \text {or}\ \mathtt{bab}\ \text {as a proper substring}]$

By definition,

$$\begin{aligned} B_5(t) \le \sum _{i \ge 7} B_5^i (t) \, \end{aligned}$$

Since there are two $3$-overlaps in bad blocks of length $6$, $2 \chi _{=6} = \#_{{\text {bad}}}^6$.

Consider bad blocks of length $i \ge 7$. Each such block contains $i-4$ $3$-overlaps. Note that overlaps $\mathtt{aba}$ or $\mathtt{bab}$ that are counted in $B_5^i (\mathtt{aba})$ cannot be neighbouring as otherwise $B_5^i$ would contain blocks $\mathtt{ababa}$ and $\mathtt{babab}$ (while this is only possible if the initial set $\mathcal S$ contains equal strings).

Let $\mathtt{aba}$ be the first overlap in a block. Then this block has prefix $\mathtt{cab}$ for $\mathtt{c} \in \Sigma$. Its suffix also equals $\mathtt{cab}$ since this is a bad block. But in this case $B_2 (\mathtt{aba}) > 0$ and then $B_5 (\mathtt{aba}) = 0$, a contradiction. A similar contradiction arises if $\mathtt{aba}$ is the last overlap in a block. Thus, for $i \ge 7$ we have:

$$\begin{aligned} B_5^i = \sum _{s} B_5^i(s) \le \chi _{>5}\cdot \left\lceil \frac{i - 6}{2} \right\rceil \le \chi _{>5}\cdot (i - 6) \, . \end{aligned}$$

Then

$$\begin{aligned} 2 \chi _{>5}^i + B_5^i \le 2 \chi _{>5}+ \chi _{>5}\cdot (i - 6) = \chi _{>5}\cdot (i - 4) \le \#_{{\text {bad}}}^i \, . \end{aligned}$$

Finally, we have:

$$\begin{aligned} 2 \chi _{>5}+ \chi _{=5}+ B_5 \le \chi _{=5}+ 2 \chi _{=6} + \sum _{i \ge 7}&(2 \chi _{=i} + B_5^i) \\ \nonumber&\le \#_{{\text {bad}}}^5 + \#_{{\text {bad}}}^6 + \sum _{i \ge 7} \#_{{\text {bad}}}^i = \#_{{\text {bad}}}\, . \end{aligned}$$

   $\square$

4 Conclusion

We have proved that the greedy conjecture for the shortest common superstring problem is true for strings of length $4$. Extending the proof to the case of $5$-strings seems to be even more tedious. At the same time resolving such special cases does not seem to help to resolve the general case.